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Ymorist [56]
3 years ago
13

K2SO4 H3PO4 and NaOH KOH and H2SO4 HPO4 and KOH

Chemistry
2 answers:
Andrej [43]3 years ago
7 0

Answer: b

Explanation:

mylen [45]3 years ago
3 0

Answer: B and the answer for the second question is A

Explanation:

I just took the assignment hope this helps:)

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Considering 0.10 m solutions of each substance, which contains the smallest concentration of ions
mihalych1998 [28]

Question options:

A) K2SO4

B) FeCl₃

C) NaOH

D) NH₃

E) KCl

Answer:

D. NH₃

Explanation:

K2SO4 = 2 K+ + SO42-

[K+]= 2 x 1.0 = 2.0 M ; [SO42-] = 1.0 M

total concentrations of ions = 2.0 + 1.0 = <em>3.0 M</em>

FeCl3 = Fe3+ + 3Cl-

[Fe3+] = 1.0 M ; [Cl-] = 3 x 1.0 = 3.0

total concentration ions = 1.0 + 3.0 =<em> 4.0 M</em>

NaOH = Na+ + OH-

[Na+] = [OH-] = 1.0 M

total concentration ions = 1.0 + 1.0 = <em>2.0 M</em>

<u>NH3 is a weak acid so the concentration of NH4+ and OH- </u><u><em>< 2.0</em></u>

KCl = K+ + Cl-

[K+] = [Cl-] = 1.0 M

total concentration ions = 1.0 + 1.0 =<em> 2.0 M</em>

7 0
3 years ago
23°C = ______ K<br><br> -250<br> 296<br> 123<br> 23
strojnjashka [21]

Answer:

296

Explanation:

6 0
3 years ago
Read 2 more answers
Which equation is balanced?
vichka [17]

Answer:

2Na+F2 yields 2NaF is balanced.

Explanation:

There are 2 sodium and 2 fluorine in both reactants and product: In 2NaF the 2 is distributed because it is in the beginning of the compound.

7 0
3 years ago
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Blade resistance hair motion pull down
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3 years ago
The heat capacity of water is 1cal degree'g1 (1 calorie per degree centigrade, per gram). You are given 1 gallon of water at 25
sdas [7]

Answer:

The heat needed to boil 1 gallon of water is 81,490.62 Joules.

Explanation:

Q=mc\Delta T

Where:

Q = heat absorbed or heat lost

c = specific heat of substance

m = Mass of the substance  

ΔT = change in temperature of the substance

We have :

Volume of water = V = 1 gal = 4546.09 mL

Density of water , d= 1 g/mL

mass of water = m = d × V = 1g/mL × 4546.09 mL =  4546.09 g

Specific heat of water = c = 1 Cal/g°C

ΔT = 100°C - 25°C = 75 °C

9 (boiling pint of water is 100°C)

Heat absorbed by the water to make it boil:

Q= 4546.09 g\times 1 Cal/g^oC\times 75^oC=340,956.75 Cal

1 calorie = 4.184 J

Q=\frac{340,956.75}{4.184} J = 81,490.62 J

The heat needed to boil 1 gallon of water is 81,490.62 Joules.

5 0
3 years ago
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